Evaluate $~~\int^\pi_{\pi/2} \big(-x\cos x\big)\,dx\,$. Choose 1 answer: Choose 1 answer: (Choice A) A $1-\dfrac\pi2$ (Choice B) B $\dfrac{3\pi}2-1$ (Choice C) C $1+\dfrac\pi2$ (Choice D) D $\dfrac{3\pi}2$
Explanation: We will solve this by integrating by parts. We know that $ \int u(x)v~^\prime(x)dx = u(x)v(x)-\int u~^\prime(x)v(x)dx\,$. We can rewrite this as $ \int u\ dv = uv-\int v\ du\,$. In this problem we will let $~u = x~$ and $~dv=\cos x\, dx\,$. Then $~du = dx~$ and $~v = \sin x\,$. Integration by parts gives $ -\int^\pi_{\pi/2} x\cos x\,dx = -\bigg(x\sin x-\int\sin x\,dx\bigg)\Bigg|^\pi_{\pi/2}$ $ ~~=-\big(x\sin x+\cos x\big)\bigg]^\pi_{\pi/2}$ $ ~~ = -x\sin x-\cos x\bigg]^\pi_{\pi/2}$ $ ~~=\Big(-\pi\sin\pi-\cos\pi\Big)-\Big(-\dfrac\pi2\sin\dfrac\pi2-\cos\dfrac\pi2\Big)$ $ ~~=\big(0-(-1)\big)-\Big(-\dfrac\pi2-0\Big)= 1+\dfrac\pi2\,$.